Find the limits. Use L'Hospital as needed.

Question

limit as x approaches 0 of (cscx-cotx)

I started off by combining the two, since they have the common denomonator of sinx, so (1-cosx)/sinx. Since I got type 0/0 for this equation, I differentiated as the ratios (as f(x) and g(x)), but I wasn't sure how to finish.

limit as x approaches 0 from the right of (tan2x)^x

For this one, do I use the indeterminate powers? So, since it's type 0⁰, I would take the natural log so lny=xln(tan2x).

Answer 1

limit as x approaches 0 of (cscx-cotx)

For this part, rewrite cscx as 1/(sinx) and cotx as cosx/sinx. Now when you subtract, you get (1-cosx)/sinx. Plugging in 0 for x, you get (1-1)/0 = 0/0, so we need to use L’Hospital to take the derivative of the numerator and denominator.

The derivative of 1-cosx is the derivative of 1 minus the derivative of cosx. The derivative of 1 is 0 and the derivative of cosx is -sinx. Now we subtract. 0-(-sinx) = sinx.

Now let’s handle the denominator. The derivative of sinx is cosx. So after applying L’Hospitals Rule, we have the limit as x approaches 0 of sinx/cosx, which equals 0/1 = 0.

limit as x approaches 0 from the right of (tan2x)^x

For this one, the steps you outlined in your question are correct! Following from that, we have lny=xln(tan(2x)) and can take the limit as x approaches 0 from the right of both sides.

This next step is tricky. But remember, L’Hospitals Rule involves fractions. So let’s see if we can write xln(tan(2x)) as a fraction. In fact, we can write it as ln(tan(2x))/(1/x).

Taking the derivative of the numerator, we have to apply the derivative rule for natural logs, followed by a couple of chain rules. First we have 1/(tan(2x)). Then, applying the chain rule, we differentiate tan(2x) and get sec²(2x). Chain rule when differentiating tan(2x) means we take the derivative of 2x, which is 2. Putting it all together, the derivative of ln(tan(2x)) is 2sec²(2x)/tan(2x).

For the denominator, the derivative of (1/x) is -1/x². Now let’s simplify the differentiated numerator and denominator. We have -2x²sec²(2x)/tan(2x). And we can rewrite sec²(2x) as 1/cos²(2x) and tan(2x) as sin(2x)/cos(2x). Which leads to -2x²cos(2x)/(sin(2x)cos²(2x)). A cos(2x) cancels out and we are left with -2x²/(sin(2x)cos(2x)).

We still can’t solve this easily, because plugging in 0 yields an indeterminate answer. Let’s rewrite this into something that we can apply L’Hospitals Rule to and will be not too difficult to differentiate. We can do [2x/sin(2x)][-x/cos(2x)]. We see that 2x/sin(2x) is indeterminate, so let’s differentiate the numerator and denominator. We get 2/(2cos(2x)). And plugging in 0, we get 2/2 = 1. Now let’s plug in 0 to -x/cos(2x). This is going to be 0. And 1 times 0 is 0.

Now we’re getting somewhere! What does this 0 tell us? In an earlier step, we had lny=xln(tan(2x)) and took the limit as x approaches 0 from the right. So this 0 we just found in our last step tells us that the limit as x approaches 0 from the right of lny is equal to 0. But we want the limit of y. In order to isolate the y, we can use e. Applying this, we have y = e^lny. So the limit as x approaches 0 from the right of y is equal to the limit as x approaches 0 from the right of e^lny. And we have the limit for lny, which is 0. Finally, the limit as x approaches 0 from the right of y is equal to e⁰, which is 1.

Let me know if you have any questions on either of these problems! The second one is certainly a bit of a challenge, so I’d be happy to clarify anything further if you need.